Q.A plane electromagnetic wave of frequency $25\ \text{MHz}$ travels in free space along the $x$-direction. At a particular point in space and time, $\mathbf{E} = 6.3\ \hat{j}\ \text{V/m}$. What is $\mathbf{B}$ at this point?

Concept: Electromagnetic Wave Relation — in free space, $\mathbf{E}$ and $\mathbf{B}$ are perpendicular, in phase, and related by $c = E/B$.

Step 1: The wave travels along $x$, and $\mathbf{E}$ is along $\hat{j}$. For a plane wave, $\mathbf{B}$ must be perpendicular to both the direction of propagation and $\mathbf{E}$, so $\mathbf{B}$ is along $\hat{k}$.

Step 2: The magnitude relation is $B = E / c$, where $c = 3 \times 10^8\ \text{m/s}$.

Step 3:

Step 4: The direction is $\hat{k}$, so $\mathbf{B} = 2.1 \times 10^{-8}\ \hat{k}\ \text{T}$.

The key idea is that in a plane electromagnetic wave, the electric and magnetic fields are not independent — they are linked by Maxwell’s equations. For a wave traveling in free space, the ratio of their magnitudes is fixed by the speed of light, and their directions are perpendicular to each other and to the direction of propagation.

The wave moves along the $x$-direction. The electric field is given as $\mathbf{E} = 6.3\ \hat{j}\ \text{V/m}$, which points along the $y$-axis. For the wave to travel along $x$, the magnetic field must lie along the $z$-axis — that’s the only remaining perpendicular direction. The sign (whether $+\hat{k}$ or $-\hat{k}$) is determined by the fact that $\mathbf{E} \times \mathbf{B}$ must point in the direction of wave travel, which is $+\hat{i}$.

Let’s work through it step by step.

1. Recall the fundamental relation In free space, the magnitudes of $\mathbf{E}$ and $\mathbf{B}$ in an electromagnetic wave satisfy

where $c = 3 \times 10^8\ \text{m/s}$ is the speed of light. This comes directly from Maxwell’s equations — the wave equation for $\mathbf{E}$ and $\mathbf{B}$ gives the same speed $c$, and the fields are in phase with this ratio.

2. Find the magnitude of $\mathbf{B}$ Given $E = 6.3\ \text{V/m}$, we have

3. Determine the direction The wave travels along $+\hat{i}$. The electric field is along $+\hat{j}$. For the Poynting vector $\mathbf{S} = \frac{1}{\mu_0} (\mathbf{E} \times \mathbf{B})$ to point along $+\hat{i}$, we need $\mathbf{E} \times \mathbf{B}$ to be along $+\hat{i}$. Using the right-hand rule: $\hat{j} \times \hat{k} = \hat{i}$. So $\mathbf{B}$ must be along $+\hat{k}$.

4. Write the final vector Therefore,

Method: The Right-Hand Rule and the Wave Relation for EM Waves

This problem uses the plane wave relation between electric and magnetic fields in free space, combined with the direction rule for electromagnetic waves.

Key Concept

For a plane EM wave traveling in free space:

• $\mathbf{E}$, $\mathbf{B}$, and the direction of propagation $\hat{k}$ are mutually perpendicular.
• The magnitudes are related by:

where $c = 3 \times 10^8\ \text{m/s}$.

Steps

1. Identify the direction of propagation

   The wave travels along the $x$-direction. So $\hat{k} = \hat{i}$.

2. Identify the direction of $\mathbf{E}$

   Given: $\mathbf{E} = 6.3\ \hat{j}\ \text{V/m}$. So $\mathbf{E}$ points along $+y$.

3. Apply the right-hand rule

   For a wave traveling in the $+\hat{k}$ direction:

Here $\hat{k} = \hat{i}$, $\hat{E} = \hat{j}$.

Using $\hat{i} = \hat{j} \times \hat{B}$, we get $\hat{B} = \hat{k}$ (the $+z$ direction).

4. Calculate the magnitude of $\mathbf{B}$

5. Write the final vector

Quick Check

• Frequency $25\ \text{MHz}$ is not needed here — it only confirms the wave is in the radio band, but the relation $E = cB$ is frequency-independent in free space.
• The direction matches: $x$-propagation, $y$-electric field, $z$-magnetic field.

This is a classic problem from the Electromagnetic Waves chapter in NCERT Class 12 Physics. Here's a breakdown of the common mistakes students make on it, and how to avoid them.

🔍 The Correct Approach First

For an EM wave in free space:

• $\mathbf{E}$, $\mathbf{B}$, and direction of propagation are mutually perpendicular.
• Relation: $|\mathbf{B}| = \frac{|\mathbf{E}|}{c}$, where $c = 3 \times 10^8\ \text{m/s}$.
• Direction: $\mathbf{E} \times \mathbf{B}$ gives the direction of wave travel.

Here:

• Wave travels along $+x$.
• $\mathbf{E} = 6.3\ \hat{j}\ \text{V/m}$ (along $+y$).
• So $\mathbf{B}$ must be along $+z$ (since $\hat{j} \times \hat{k} = \hat{i}$).

Calculation:

Final answer:

✗ Common Mistake #1: Forgetting the Direction Rule

What students do wrong:

They calculate magnitude correctly but write $\mathbf{B}$ along $+y$ or $+x$, or just give magnitude.

Why it happens:

They memorise "E and B are perpendicular" but don't apply the right-hand rule or the cross-product relation $\mathbf{E} \times \mathbf{B} \parallel \text{propagation direction}$.

How to avoid:

• Always write: propagation direction = $\mathbf{E} \times \mathbf{B}$ direction.
• Use unit vectors: $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, etc.
• Practice with all three axes.

✗ Common Mistake #2: Using Wrong Value of $c$

What students do wrong:

They use $c = 3 \times 10^8\ \text{m/s}$ but sometimes mistakenly use $3 \times 10^8\ \text{km/s}$ or forget the exponent.

Why it happens:

Rushing or not writing the formula clearly.

How to avoid:

• Always write $c = 3 \times 10^8\ \text{m/s}$ at the top.
• Double-check units: E in V/m, B in T.
• If frequency is given, it's a distractor — you don't need it here.

✗ Common Mistake #3: Using Frequency Unnecessarily

What students do wrong:

They try to use $c = f\lambda$ or $B = \frac{E}{c}$ with frequency, leading to wrong numbers.

Why it happens:

The problem gives frequency (25 MHz) — students think it must be used.

How to avoid:

• Recognise: For a plane wave in free space, $B = E/c$ is always true, independent of frequency.
• Frequency is only needed if they ask for wavelength or wave number.

✗ Common Mistake #4: Unit Confusion

What students do wrong:

They write B in Gauss instead of Tesla, or forget to convert MHz.

Why it happens:

Mixing CGS and SI units.

How to avoid:

• Stick to SI: E in V/m, B in T, c in m/s.
• 1 T = $10^4$ G — but NCERT uses Tesla.
• Frequency in Hz: 25 MHz = $25 \times 10^6$ Hz (but again, not needed here).

✓ Quick Checklist to Avoid All Mistakes

Final takeaway:

In free space, $B = E/c$ always. The direction is the only tricky part — use the cross-product rule carefully.